n8fau

Hello all.....

I've used NPN transistors as switches for years and would like to try using a power mosfet as a switch.

How do you calculate the resistor that would go in the gate circuit to fully turn on the mosfet?

Next question is how to select the right device to do the job?

Thanks in advance.....

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73
Bob
n8fau

n8fau

I just figured out how to attach my circuit diagram....


See attached

Attached Files

mosfet.pdf (27.8 KB, 39 views)

__________________
73
Bob
n8fau

cdstahl

the device turns on based on gate voltage. some times a small series resistor is added in series with the gate for other reasons. the gate acts as a fairly large capacitor, and sometimes the gate is driven by an inductive source, in which case there can be enough ringing to cause the gate to turn off accidently. a small series resistor is added to damp this out.

otherwise, there is nearly no need for a series resistor.

further, you no longer need feedback to keep the device from saturating. the FET will still turn off very fast. often, the threshold may be 5V, but the gate will be driven to 10V and back to 0V. this mainly causes a delay -- the gate drops to around 5V, then the device turns off, then the gate drops to 0V.

so, basically, just apply a high voltage (above the threshold) from a low impedance source and your done.

The FET will stay on until driven low -- the gate capacitance is high enough and near no current flows through the gate. so in some cases, you'll need a pulldown.

Also, transconductane is lower, but this usually isn't an issue. but the required Vgs to "saturate" the device may be a bit higher then the threshold. but this does make the "emitter follower" equivilent fairly poor, as the gate may need to go 5-10V above the source. but methods exist to allow this to work for rapidly switched circuits.

lastlt, "saturate" is backwards in FETs -- the physists decided that "saturate" would be the point at which current is "saturated", which makes the "saturated mode" of the FET similar to the "linear mode" of a BJT... likewise "linear mode" for FETs is very similar to the "saturated" BJT. its somewhat annoying if you aren't aware of the terminology differences.

Diver300

To get the N-type MOSFET to turn on fully, the gate has to be at least the threshold voltage above the source voltage.

One of your circuits has the load connected to ground. If you want to turn the load on completely, so that there is very little voltage drop, you will need to have a voltage above the power supply to turn on the gate.

If you want a positive supplied load connected to ground, you would be better to use a P-type MOSFET

audioguru

No.
The current in a Mosfet with the gate at its threhold voltage is a very low current, 0.25mA. It is the threshold where it just begins to turn on. Most Mosfets need a gate to source voltage of 10V to fully turn on and their threshold voltage is 2V to 4V.

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Uncle $crooge

Hero999

Look at the graph on the datasheet which shows the drain current, Vds and the gate voltage.

For example the ultra cheap 2N7000 is fine for driving <75mA loads when the gate voltage is only 4.5V but if you try to drive a 200mA load when Vgs is only 4.5V the power dissipation of the MOSFET will be exceeded causing lots of smoke.
http://www.fairchildsemi.com/ds/2N%2F2N7000.pdf

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Bob Scott

Which makes the 2N7000 perfect for driving LEDs from 5V CMOS logic.

Bob